JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 13)
An alternating emf $$\mathrm{E}=440 \sin 100 \pi \mathrm{t}$$ is applied to a circuit containing an inductance of $$\frac{\sqrt{2}}{\pi} \mathrm{H}$$. If an a.c. ammeter is connected in the circuit, its reading will be :
4.4 A
1.55 A
2.2 A
3.11 A
คำอธิบาย
$$I = {V \over {\omega L}}$$
$$ = {{440} \over {100\pi \times {{\sqrt 2 } \over \pi }}} = {{44} \over {10\sqrt 2 }}$$
$$ \Rightarrow {I_{rms}} = {I \over {\sqrt 2 }} = {{44} \over {20}} = 2.2\,A$$